How to print string from function which returns pointer on that string in C?

Question

This is my program in C:

#include <stdio.h>

char const* voice(void){
  return "hey!";
}

int main(){
const char* (*pointer)(void);
pointer = &voice;


printf ("%s\n", *pointer); // check down *
return 0;
}

*with this i'm trying to print what is returning from pointer but it seems like not working.

What am I doing wrong?

I guess you want to keep the function pointer stuff, so just call the function instead of dereferencing; printf( "%s\n", pointer( ) ); — Dave
– Dave, Commented Apr 2, 2013 at 1:34

Pubby · Accepted Answer · 2013-04-02 01:32:19Z

6

You need to call the function pointers, ie use parenthesis:

#include <stdio.h>

char const* voice(void){
  return "hey!";
}

int main(){
const char* (*pointer)(void);
pointer = &voice;


printf ("%s\n", pointer());
//              ^^^^^^^^^
return 0;
}

* isn't needed for function pointers. (neither is &)

answered Apr 2, 2013 at 1:32

Pubby

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1 Comment

Tommz Over a year ago

Thanks a lot. Yea, I was asking myself why pointer = voice; is working as well.

Sergey Kalinichenko · Accepted Answer · 2013-04-02 01:32:15Z

4

You call a function through a pointer in the same exact way that you call a function directly, as if that pointer was function's name:

printf ("%s\n", pointer());

Starting with the ANSI C standard, the asterisks and ampersands around function pointers are optional.

answered Apr 2, 2013 at 1:32

Sergey Kalinichenko

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1 Comment

Jonathan Leffler Over a year ago

The alternative notation is (*pointer)(), with parentheses and an asterisk. *pointer() is valid but returns the first character of the string returned by the function pointer.

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